028.19. a) (12 \frac{12}{13})^2; б) (14 \frac{13}{15})^2; B) (39 \frac{39}{40})^2; r) (15 \frac{13}{16})^2

Решение:

Используем формулу квадрата суммы $$ (a+b)^2 = a^2 + 2ab + b^2 $$.

• 028.19. a)

  $$ \left(12\frac{12}{13}\right)^2 = \left(\frac{12\cdot13+12}{13}\right)^2 = \left(\frac{156+12}{13}\right)^2 = \left(\frac{168}{13}\right)^2 = \frac{168^2}{13^2} = \frac{28224}{169} $$

• 028.19. б)

  $$ \left(14\frac{13}{15}\right)^2 = \left(\frac{14\cdot15+13}{15}\right)^2 = \left(\frac{210+13}{15}\right)^2 = \left(\frac{223}{15}\right)^2 = \frac{223^2}{15^2} = \frac{49729}{225} $$

• 028.19. B)

  $$ \left(39\frac{39}{40}\right)^2 = \left(\frac{39\cdot40+39}{40}\right)^2 = \left(\frac{1560+39}{40}\right)^2 = \left(\frac{1599}{40}\right)^2 = \frac{1599^2}{40^2} = \frac{2556801}{1600} $$

• 028.19. r)

  $$ \left(15\frac{13}{16}\right)^2 = \left(\frac{15\cdot16+13}{16}\right)^2 = \left(\frac{240+13}{16}\right)^2 = \left(\frac{253}{16}\right)^2 = \frac{253^2}{16^2} = \frac{64009}{256} $$

Ответ: 28224/169; 49729/225; 2556801/1600; 64009/256.